SOLUTION: how to solve radical xsquare minus 4x minus 12 minus radical xsquare minus 4x minus 5 =1

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Question 250500: how to solve radical xsquare minus 4x minus 12 minus radical xsquare minus 4x minus 5 =1
Answer by jsmallt9(3758)   (Show Source): You can put this solution on YOUR website!

A procedure to solve square root equations:
  1. Isolate a square root.
  2. Square both sides of the equation.
  3. If there is still a square root, repeat steps 1 and 2.
  4. Solve the equation (which should no longer have any square roots).
  5. Check the solutions. This is more than just a good idea. It is important. Squaring both sides of an equation (Step #2) may introduce what are called extraneous solutions. Extraneous solutions are solutions which fit the squared equation but do not fit the original equation. Extraneous solutions, if any, are found by checking solutions in the original equation. Extraneous solutions must be rejected.

Let's see how this works.
1. Isolate a square root.
Add to both sides:


2. Square both sides

The right side is the square of a binomial (2-term expression). Be sure to use FOIL or the binomial square pattern, to square it correctly.



3. Repeat steps 1 and 2 (since we still have a square root.
Subtract from, add 4x to and add 4 to each side:

Divide both sides by 2:

(At this point, if you're clever, you will realize that there will be no solutions to this equation. Square roots are positive so there is no way for one to be equal to -4. But we will continue as if we didn't notice this.)



4. Solve the equation.
Since this equation is quadratic, we will make one side equal to zero by subtracting 16 from each side:

This factors pretty easily. (You can use the Quadratic Formula if you prefer.)

By the Zero Product Property this product can only be zero if one of the factors is zero. So:
or
Solving these we get:
or

5.Check the solution using the original solution:

Checking x = 7:





Does not check. Reject x = 7.
Checking x = -3:





Does not check. Reject x = -3.
Neither of the two solutions found actually work in the original equation. So there is no solution to the original equation.
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